Answer:hello the answer is C. 7$
Step-by-step explanation:
A.)
<span>s= 30m
u = ? ( initial velocity of the object )
a = 9.81 m/s^2 ( accn of free fall )
t = 1.5 s
s = ut + 1/2 at^2
\[u = \frac{ S - 1/2 a t^2 }{ t }\]
\[u = \frac{ 30 - ( 0.5 \times 9.81 \times 1.5^2) }{ 1.5 } \]
\[u = 12.6 m/s\]
</span>
b.)
<span>s = ut + 1/2 a t^2
u = 0 ,
s = 1/2 a t^2
\[s = \frac{ 1 }{ 2 } \times a \times t ^{2}\]
\[s = \frac{ 1 }{ 2 } \times 9.81 \times \left( \frac{ 12.6 }{ 9.81 } \right)^{2}\]
\[s = 8.0917...\]
\[therfore total distance = 8.0917 + 30 = 38.0917.. = 38.1 m \] </span>
Answer:
(-3,-20)
Step-by-step explanation:
4 +8x = y ...... equation (1)
6x - y = 2 ........equation(2)
from equation 2
6x - y = 2
6x -(4+8x) =2
6x - 4 + 8x = 2
+4 +4
6x - 8x = 6
-2x=6
divide by -2
x= -3
put x in equation (1) to solve for y
4 + 8x = y
4+8(-3)=y
4-24=y
-20=y
y= -20
solution = (-3,-20)
Answer: I'm guessing it would be a parabola, with the line going through -6 on the y-axis and passing through 2 and 6 on the x-axis, but we cannot see any answers, so therefore we can't answer it accurately.
Step-by-step explanation:
